The distance between the plates of a parallel plate capacitor is $5d$. The positively charged plate is at $x=0$ and the negatively charged plate is at $x=5d$. Two slabs,one of a conductor and the other of a dielectric,both of equal thickness $d$,are inserted between the plates as shown in the figure. The potential versus distance graph will look like:

In the circuit shown below,the switch $S$ is connected to position $P$ for a long time so that the charge on the capacitor becomes $q_1 \mu C$. Then $S$ is switched to position $Q$. After a long time,the charge on the capacitor is $q_2 \mu C$.
$(1)$ The magnitude of $q_1$ is
$(2)$ The magnitude of $q_2$ is
Give the answer of question $(1)$ and $(2)$

For the given circuit,find the potential at point $a$. (in $V$)

$A$ capacitor of capacitance $C$ is initially charged to a potential difference of $V$ $volt$. Now it is connected to a battery of $2V$ with opposite polarity. The ratio of heat generated to the final energy stored in the capacitor will be

Capacitor $A$ has a capacitance of $1\ \mu F$. It is filled with a dielectric medium of constant $K = 15$. Another capacitor $B$ has a capacitance of $1\ \mu F$. Both are charged separately by $100\ V$ batteries. After charging,the batteries are removed,the dielectric is removed from $A$,and the capacitors are connected in parallel. Calculate the common potential in $V$.

Two capacitors of capacitances $1\ \mu F$ and $3\ \mu F$ are charged to the same voltages $5\ V$. They are connected in parallel with oppositely charged plates connected together. Then: